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An adult weighting $600 \mathrm{~N}$ raises the centre of gravity of his body by $0.25 \mathrm{~m}$ while taking each step of $1 \mathrm{~m}$ length in jogging. If he jogs for $6 \mathrm{~km}$, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting $10 \%$ of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilized for jogging.
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Verified Answer
As given that,
Weight of the adult $(W)=m g=600 \mathrm{~N}$
Height of each step $=h=0.25 \mathrm{~m}$
Length of each step $=1 \mathrm{~m}$
Total distance travelled $=6 \mathrm{~km}=6000 \mathrm{~m}$
So, total number of steps in $\mathrm{cm}$.
$$
=\frac{6000}{1}=6000 \text { steps. }
$$
Total energy utilised in $6000 \mathrm{~m}$ for jogging $=n \times m g h=6000 \times 600 \times 0.25 \mathrm{~J}=9 \times 10^5 \mathrm{~J}$
Since, $10 \%$ of energy is utilised in jogging
$$
\begin{aligned}
&=\left[\frac{10}{100} \times 600 \times 600 \times 0.25\right] \\
&=360000 \times 0.25=90000 \mathrm{~J}=9 \times 10^4 \mathrm{~J}
\end{aligned}
$$
Weight of the adult $(W)=m g=600 \mathrm{~N}$
Height of each step $=h=0.25 \mathrm{~m}$
Length of each step $=1 \mathrm{~m}$
Total distance travelled $=6 \mathrm{~km}=6000 \mathrm{~m}$
So, total number of steps in $\mathrm{cm}$.
$$
=\frac{6000}{1}=6000 \text { steps. }
$$
Total energy utilised in $6000 \mathrm{~m}$ for jogging $=n \times m g h=6000 \times 600 \times 0.25 \mathrm{~J}=9 \times 10^5 \mathrm{~J}$
Since, $10 \%$ of energy is utilised in jogging
$$
\begin{aligned}
&=\left[\frac{10}{100} \times 600 \times 600 \times 0.25\right] \\
&=360000 \times 0.25=90000 \mathrm{~J}=9 \times 10^4 \mathrm{~J}
\end{aligned}
$$
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